Worked examples · Sheet 3

Relative positions of lines and planes

Theory used: Relative position of a line and a plane Coplanar lines and skew lines Coplanarity criterion for two lines Classification of two lines

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine the relative position of the line $r : x = 1 + 2 t$ , $y = - 1 + t$ , $z = 3 t$ with respect to each of the planes

π_{1} : 2 x - y + z = 9, π_{2} : x + y - z = 2.

v = (2, 1, 3), A = (1, - 1, 0)

n_{1} \cdot v = (2, - 1, 1) \cdot (2, 1, 3) = 4 - 1 + 3 = 6 \neq = 0

2 (1 + 2 t) - (- 1 + t) + 3 t = 9 ⟹ 3 + 6 t = 9 ⟹ t = 1 ⟹ r \cap π_{1} = {(3, 0, 3)}

n_{2} \cdot v = (1, 1, - 1) \cdot (2, 1, 3) = 2 + 1 - 3 = 0, 1 + (- 1) - 0 = 0 \neq = 2

The relative position of a line and a plane is decided by the dot product of the direction vector of the line and the normal vector of the plane. For $π_{1}$ the dot product equals $6$ , so the line and the plane are intersecting at a single point. To find it, substitute the parametric equations into the equation of the plane; this gives $t = 1$ and the point $(3, 0, 3)$ , which satisfies the equation of $π_{1}$ since $6 - 0 + 3 = 9$ . For $π_{2}$ the dot product is zero, therefore the line is parallel to the plane or contained in it. The point $A$ of $r$ gives $1 - 1 - 0 = 0$ , a value different from the constant term $2$ , hence $A \in / π_{2}$ and the line is parallel to $π_{2}$ and has no points in common with it.

Exercise 2

Determine the relative position of the lines

r : ⎩ ⎨ ⎧ x = 1 + t y = t z = 2 - t s : ⎩ ⎨ ⎧ x = 1 + u y = 2 - u z = - 1 + 2 u

and, if they are intersecting, find the common point.

A = (1, 0, 2), v = (1, 1, - 1), B = (1, 2, - 1), w = (1, - 1, 2)

A B = (0, 2, - 3), v \times w = i 11 j 1 - 1 k - 1 2 = (1, - 3, - 2)

A B \cdot (v \times w) = 0 - 6 + 6 = 0

⎩ ⎨ ⎧ 1 + t = 1 + u t = 2 - u 2 - t = - 1 + 2 u ⟹ t = u = 1 ⟹ r \cap s = {(2, 1, 1)}

The direction vectors $(1, 1, - 1)$ and $(1, - 1, 2)$ are not proportional, so the lines are not parallel. By the coplanarity criterion for two lines we compute the scalar triple product $A B \cdot (v \times w)$ , which turns out to be zero; therefore the lines are coplanar, and since they are not parallel they are intersecting. The common point is found by equating the coordinates of the generic points of the two lines. The first equation gives $t = u$ , the second becomes $t = 2 - t$ , hence $t = u = 1$ ; the third equation is satisfied, since $2 - 1 = - 1 + 2$ . Substituting $t = 1$ into the equations of $r$ gives the point $(2, 1, 1)$ .

Exercise 3

Determine the relative position of the lines

r : ⎩ ⎨ ⎧ x = 1 + t y = 2 t z = 0 s : {x - z = 0 y = 2

w = (1, 0, - 1) \times (0, 1, 0) = i 10 j 01 k - 1 0 = (1, 0, 1), B = (0, 2, 0) \in s

A = (1, 0, 0), v = (1, 2, 0), A B = (- 1, 2, 0)

A B \cdot (v \times w) = - 1 11 220001 = 1 \cdot - 1 1 22 = - 2 - 2 = - 4 \neq = 0

The line $s$ is given in Cartesian form, so first we compute a direction vector and a point of it. The direction vector is the cross product of the normal vectors, $w = (1, 0, 1)$ ; a point is obtained by setting $x = 0$ , which gives $z = 0$ and $y = 2$ . The direction vectors $(1, 2, 0)$ and $(1, 0, 1)$ are not parallel. The scalar triple product is computed as the determinant of the matrix with rows $A B$ , $v$ , $w$ , expanded along the third column, which has a single nonzero entry. The determinant equals $- 4 \neq = 0$ , therefore by the coplanarity criterion for two lines the lines are skew.

Exercise 4

Determine for which values of $k \in R$ the lines

r : ⎩ ⎨ ⎧ x = 1 + t y = t z = 1 s : ⎩ ⎨ ⎧ x = u y = 1 z = k + u

are coplanar and, for those values, determine their relative position.

A = (1, 0, 1), v = (1, 1, 0), B = (0, 1, k), w = (1, 0, 1)

A B \cdot (v \times w) = - 1 11 110 k - 1 01 = - 1 \cdot 1 - 1 \cdot 1 + (k - 1) \cdot (- 1) = - 1 - k

- 1 - k = 0 ⟺ k = - 1

k = - 1 : ⎩ ⎨ ⎧ 1 + t = u t = 1 1 = - 1 + u ⟹ t = 1, u = 2 ⟹ r \cap s = {(2, 1, 1)}

The direction vectors $(1, 1, 0)$ and $(1, 0, 1)$ are not parallel for any value of $k$ , so the lines are never parallel. By the coplanarity criterion for two lines they are coplanar if and only if the scalar triple product is zero. The determinant, expanded along the first row, equals $- 1 - k$ , and vanishes only for $k = - 1$ . For $k \neq = - 1$ the lines are therefore skew. For $k = - 1$ the lines are coplanar and not parallel, therefore they are intersecting. The system obtained by equating the coordinates gives $t = 1$ from the second equation and $u = 2$ from the third, and the first is satisfied, since $1 + 1 = 2$ ; the common point is $(2, 1, 1)$ .